1. ## sequence

Prove the following: Let $(a_n)$ and $(b_n)$ be sequences of real numbers that converge to real numbers $a$ and $b$ respectively. If $a_n \leq b_n$ for all $n \in \mathbb{N}$, then $a \leq b$.

Proof: For each positive real number $\epsilon$ there exists a $N_1 \in \mathbb{N}$ so that whenever $n > N_1$, $|a_n-a| < \epsilon$. Also for each positive real number $\epsilon$ there exists a $N_2 \in \mathbb{N}$ so that whenever $n > N_2$, $|b_n-b| < \epsilon$. So $a- \epsilon < a < a + \epsilon$ and $b- \epsilon < b < b+ \epsilon$. Then $0 \leq b_n- a_n < b-a$ for all $n \in \mathbb{N}$. It follows that $a \leq b$. $\square$

Is this correct?

2. Originally Posted by manjohn12
Prove the following: Let $(a_n)$ and $(b_n)$ be sequences of real numbers that converge to real numbers $a$ and $b$ respectively. If $a_n \leq b_n$ for all $n \in \mathbb{N}$, then $a \leq b$.

Proof: For each positive real number $\epsilon$ there exists a $N_1 \in \mathbb{N}$ so that whenever $n > N_1$, $|a_n-a| < \epsilon$. Also for each positive real number $\epsilon$ there exists a $N_2 \in \mathbb{N}$ so that whenever $n > N_2$, $|b_n-b| < \epsilon$. So $a- \epsilon < a < a + \epsilon$ and $b- \epsilon < b < b+ \epsilon$. Then $0 \leq b_n- a_n < b-a$ for all $n \in \mathbb{N}$. It follows that $a \leq b$. $\square$

Is this correct?
Try proving the following easier result: let $\{ x_n\}$ be a sequence such that $x_n \geq 0$ and $x_n \to x$ then $x\geq 0$.

After you prove that consider $x_n = a_n - b_n \geq 0$.